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Limits of Sequences

Date: 02/25/2001 at 02:11:06
From: N. D. Kapoor
Subject: A question about limits

Dear Dr. Math,

We know that the limit (n -> infinity) [(1 + 1/n)^n] = e. Is the limit 
(n -> infinity) [(1 + 1/sqrt(n))^(1.5n)] also e? Also, what is the 
limit (n -> infinity) [(1 + a/n)^n], where a is not equal to 0?

Please help.
N. D. Kapoor

Date: 02/26/2001 at 22:27:50
From: Doctor Fenton
Subject: Re: A question about limits

Dear Mr. Kapoor,

Thanks for writing to Dr. Math. For questions like these, I like to 
use the fact that for a > 0 and b real,

     a^b = e^[b*ln(a)]

where ln is the natural logarithm. Then:

     (1 + 1/sqrt(n))^(1.5n) = e^[1.5*n*ln(1 + 1/sqrt(n))]

The exponent can be written as the indeterminate form:

     ln( 1 + 1/sqrt(n))

By L'Hospital's Rule, the limit of this expression as n->oo is the 
same as the limit as n->oo of:

           1            1
      ----------- * ---------
      1 + sqrt(n)   2*sqrt(n)

and the quotient is of order n as n->oo, so the expression diverges to 
-oo, which means the original expression -> 0, not e.

For the second expression, you can repeat this computation, or write:

     (1+a/n)^n = (1 + a/n)^[(n/a)*a]

               = [(1 + a/n)^(n/a)]*a

               = {[ 1 + 1/(n/a)]^(n/a)}^a 

and the term in braces {} approaches e by your earlier observation.

If you have further questions, please write us again.

- Doctor Fenton, The Math Forum
Associated Topics:
High School Calculus
High School Sequences, Series

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